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thomas5267

  • one year ago

Prove that the polynomials \(P_n\) has n distinct root for all n. \(P_n\) are characteristic polynomials of a particular type of matrix. \[ P_n=-xA_{n-1}-\frac{1}{2}A_{n-2}\\ A_n = \left( \dfrac{-x+\sqrt{x^2-1}}{2}\right)^n + \left(\dfrac{-x-\sqrt{x^2-1}}{2}\right)^n \] \(A_n\) satisfies the following recurrence relation: \[ A_n=-x A_{n-1}-\frac{1}{4}A_{n-2}\\ A_1=-x\\ A_2=x^2-\frac{1}{2} \]

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  1. anonymous
    • one year ago
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    im not sure if we need to find th characteristic coefficient and show they are distinct

  2. anonymous
    • one year ago
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    @zzr0ck3r

  3. zzr0ck3r
    • one year ago
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    no idea

  4. thomas5267
    • one year ago
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    I want to prove these matrices are diagonalisable. \[ M_5=\begin{pmatrix} 0&\frac{1}{2}&0&0&0\\ 1&0&\frac{1}{2}&0&0\\ 0&\frac{1}{2}&0&\frac{1}{2}&0\\ 0&0&\frac{1}{2}&0&1\\ 0&0&0&\frac{1}{2}&0 \end{pmatrix}\\ M_7=\begin{pmatrix} 0&\frac{1}{2}&0&0&0&0&0\\ 1&0&\frac{1}{2}&0&0&0&0\\ 0&\frac{1}{2}&0&\frac{1}{2}&0&0&0\\ 0&0&\frac{1}{2}&0&\frac{1}{2}&0&0\\ 0&0&0&\frac{1}{2}&0&\frac{1}{2}&0\\ 0&0&0&0&\frac{1}{2}&0&1\\ 0&0&0&0&0&\frac{1}{2}&0 \end{pmatrix}\\ \] In order words, \((M_n)_{i,i+1}=(M_n)_{i+1,i}=\frac{1}{2}\) except for \((M_n)_{2,1}=(M_n)_{n-1,n}=1\)

  5. thomas5267
    • one year ago
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    The old question has gotten so convoluted I figured that I will post a new one.

  6. zzr0ck3r
    • one year ago
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    This is just an algorithm...

  7. anonymous
    • one year ago
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    seems diagonalizable , but look at the diagonal zero which might have zero and its determinant might be zero in both and not being singular not sure though i can't help. maybe @Empty

  8. zzr0ck3r
    • one year ago
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    find the eigen values and the eigen vectors

  9. thomas5267
    • one year ago
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    As far as I can check, from \(M_3\) to \(M_{15}\) the matrices all have distinct eigenvalues.

  10. zzr0ck3r
    • one year ago
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    ahh you are trying to show for the whole family?

  11. thomas5267
    • one year ago
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    Yes. Favard's theorem seems to apply in here. Those polynomials should be orthogonal.

  12. anonymous
    • one year ago
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    well, there is a reason why i hate orthogonality. i'll watch and learn!

  13. anonymous
    • one year ago
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    @dan815 @Kainui @Empty when ever your free just lets try on this!

  14. thomas5267
    • one year ago
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    I know nothing about Favard's theorem but I found this seemingly useful theorem on google.

  15. thomas5267
    • one year ago
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    I have no idea whether \(A_n\) is orthogonal or not. It does not seem like the case.

  16. Empty
    • one year ago
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    Can you use the Cayley Hamilton theorem? It says that every matrix satisfies its characteristic equation.

  17. thomas5267
    • one year ago
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    What I want to prove is that the matrices \(M_n\) is always diagonalisable. A sufficient but not necessary condition is that the characteristic polynomials of \(M_n\) have n distinct roots. Not sure how Cayley Hamilton Theorem will help though.

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